Question

In 8086 microprocessor, if DS = 1100H, BX = 0200 H and SI = 0500H, the address accessed by MOV CH, [BX + SI] is

1. 00300 H
2. 11700 H
3. 0700 H
4. 01800 H

Answer 1

Correct Answer - Option 2 : 11700 H

MOV CH, [BX + SI]

After executing the above instruction, a byte moves from the address pointed by Bx + SI in data segment to CL

i.e. the Physical Address accessed will be:

DS * (10H) + BX + SI

Given DS = 1100H, BX = 0200 H and S1 = 0500H

Putting on the respective values, we get the physical address as:

= (1100 × 10) + 0200 + 0500

= 11000 + 0200 + 0500

= 11700 H